two-phase heat conductors with a stationary isothermic …two-phase heat conductors with a...

Rend. Istit. Mat. Univ. TriesteVolume 48 (2016), 167–187

DOI: 10.13137/2464-8728/13155

Two-phase heat conductors witha stationary isothermic surface

Shigeru Sakaguchi

Dedicated to Professor Giovanni Alessandrini on his sixtieth birthday

Abstract. We consider a two-phase heat conductor in RN with N ≥ 2consisting of a core and a shell with different constant conductivities.Suppose that, initially, the conductor has temperature 0 and, at alltimes, its boundary is kept at temperature 1. It is shown that, if thereis a stationary isothermic surface in the shell near the boundary, thenthe structure of the conductor must be spherical. Also, when the mediumoutside the two-phase conductor has a possibly different conductivity,we consider the Cauchy problem with N ≥ 3 and the initial conditionwhere the conductor has temperature 0 and the outside medium hastemperature 1. Then we show that almost the same proposition holdstrue.

Keywords: heat equation, diffusion equation, two-phase heat conductor, transmissioncondition, initial-boundary value problem, Cauchy problem, stationary isothermic sur-face, symmetry.MS Classification 2010: 35K05, 35K10, 35B40, 35K15, 35K20.

1. Introduction

Let Ω be a bounded C2 domain in RN (N ≥ 2) with boundary ∂Ω, and letD be a bounded C2 open set in RN which may have finitely many connectedcomponents. Assume that Ω \ D is connected and D ⊂ Ω. Denote by σ =σ(x) (x ∈ RN ) the conductivity distribution of the medium given by

σ =

σc in D,

σs in Ω \D,σm in RN \ Ω,

where σc, σs, σm are positive constants and σc 6= σs. This kind of three-phaseelectrical conductor has been dealt with in [7] in the study of neutrally coatedinclusions.

168 SHIGERU SAKAGUCHI

In the present paper we consider the heat diffusion over two-phase or three-phase heat conductors. Let u = u(x, t) be the unique bounded solution ofeither the initial-boundary value problem for the diffusion equation:

ut = div(σ∇u) in Ω× (0,+∞), (1)

u = 1 on ∂Ω× (0,+∞), (2)

u = 0 on Ω× 0, (3)

or the Cauchy problem for the diffusion equation:

ut = div(σ∇u) in RN × (0,+∞) and u = XΩc on RN × 0, (4)

where XΩc denotes the characteristic function of the set Ωc = RN \Ω. Considera bounded domain G in RN satisfying

D ⊂ G ⊂ G ⊂ Ω and dist(x, ∂Ω) ≤ dist(x,D) for every x ∈ ∂G. (5)

The purpose of the present paper is to show the following theorems.

Theorem 1.1. Let u be the solution of problem (1)-(3) for N ≥ 2, and let Γbe a connected component of ∂G satisfying

dist(Γ, ∂Ω) = dist(∂G, ∂Ω). (6)

If there exists a function a : (0,+∞)→ (0,+∞) satisfying

u(x, t) = a(t) for every (x, t) ∈ Γ× (0,+∞), (7)

then Ω and D must be concentric balls.

Corollary 1.2. Let u be the solution of problem (1)-(3) for N ≥ 2. If thereexists a function a : (0,+∞)→ (0,+∞) satisfying

u(x, t) = a(t) for every (x, t) ∈ ∂G× (0,+∞), (8)

then Ω and D must be concentric balls.

Theorem 1.3. Let u be the solution of problem (4) for N ≥ 3. Then thefollowing assertions hold:

(a) If there exists a function a : (0,+∞) → (0,+∞) satisfying (8), then Ωand D must be concentric balls.

(b) If σs = σm and there exists a function a : (0,+∞)→ (0,+∞) satisfying(7) for a connected component Γ of ∂G with (6), then Ω and D must beconcentric balls.

TWO-PHASE HEAT CONDUCTORS 169

Corollary 1.2 is just an easy by-product of Theorem 1.1. Theorem 1.3is limited to the case where N ≥ 3, which is not natural; that is required fortechnical reasons in the use of the auxiliary functions U, V,W given in section 4.We conjecture that Theorem 1.3 holds true also for N = 2.

The condition (7) means that Γ is an isothermic surface of the normalizedtemperature u at every time, and hence Γ is called a stationary isothermic sur-face of u. When D = ∅ and σ is constant on RN , a symmetry theorem similarto Theorem 1.1 or Theorem 1.3 has been proved in [13, Theorem 1.2, p. 2024]provided the conclusion is replaced by that ∂Ω must be either a sphere or theunion of two concentric spheres, and a symmetry theorem similar to Corol-lary 1.2 has also been proved in [10, Theorem 1.1, p. 932]. The present papergives a generalization of the previous results to multi-phase heat conductors.

We note that the study of the relationship between the stationary isother-mic surfaces and the symmetry of the problems has been initiated by Alessan-drini [2, 3]. Indeed, when D = ∅ and σ is constant on RN , he considered theproblem where the initial data in (3) is replaced by the general data u0 inproblem (1)-(3). Then he proved that if all the spatial isothermic surfaces of uare stationary, then either u0 − 1 is an eigenfunction of the Laplacian or Ω isa ball where u0 is radially symmetric. See also [8, 14] for this direction.

The following sections are organized as follows. In section 2, we give fourpreliminaries where the balance laws given in [9, 10] play a key role on behalfof Varadhan’s formula (see (12)) given in [15]. Section 3 is devoted to theproof of Theorem 1.1. Auxiliary functions U, V given in section 3 play a keyrole. If D is not a ball, we use the transmission condition (35) on ∂D toget a contradiction to Hopf’s boundary point lemma. In section 4, we proveTheorem 1.3 by following the proof of Theorem 1.1. Auxiliary functions U, V,Wgiven in section 4 play a key role. We notice that almost the same argumentswork as in the proof of Theorem 1.1.

2. Preliminaries for N ≥ 2

Concerning the behavior of the solutions of problem (1)-(3) and problem (4),we start with the following lemma.

Lemma 2.1. Let u be the solution of either problem (1)-(3) or problem (4). Wehave the following assertions:

(a) For every compact set K ⊂ Ω, there exist two positive constants B and bsatisfying

0 < u(x, t) < Be−bt for every (x, t) ∈ K × (0, 1].

(b) There exists a constant M > 0 satisfying

0 ≤ 1− u(x, t) ≤ min1,Mt−N2 |Ω|


for every (x, t) ∈ Ω × (0,+∞) or ∈ RN × (0,∞), where |Ω| denotes theLebesgue measure of the set Ω.

(c) For the solution u of problem (1)-(3), there exist two positive constantsC and λ satisfying

0 ≤ 1− u(x, t) ≤ Ce−λt for every (x, t) ∈ Ω× (0,+∞).

(d) For the solution u of problem (4) where N ≥ 3, there exist two positiveconstants β and L satisfying

β−1|x|2−N ≤∫ ∞

0

(1− u(x, t)) dt ≤ β|x|2−N if |x| ≥ L,

where Ω ⊂ BL(0) = x ∈ RN : |x| < L.

Proof. We make use of the Gaussian bounds for the fundamental solutions ofparabolic equations due to Aronson [4, Theorem 1, p. 891] (see also [5, p. 328]).Let g = g(x, t; ξ, τ) be the fundamental solution of ut = div(σ∇u). Then thereexist two positive constants α and M such that

M−1(t− τ)−N2 e−

α|x−ξ|2t−τ ≤ g(x, t; ξ, τ) ≤M(t− τ)−

N2 e−

|x−ξ|2α(t−τ) (9)

for all (x, t), (ξ, τ) ∈ RN × (0,+∞) with t > τ .For the solution u of problem (4), 1− u is regarded as the unique bounded

solution of the Cauchy problem for the diffusion equation with initial dataXΩ which is greater than or equal to the corresponding solution of the initial-boundary value problem for the diffusion equation under the homogeneousDirichlet boundary condition by the comparison principle. Hence we havefrom (9)

1− u(x, t) =

∫RN

g(x, t; ξ, 0)XΩ(ξ) dξ ≤Mt−N2 |Ω|.

The inequalities 0 ≤ 1 − u ≤ 1 follow from the comparison principle. Thiscompletes the proof of (b). Moreover, (d) follows from (9) as is noted in [4, 5.Remark, pp. 895–896].

For (a), let K be a compact set contained in Ω. We set

Nρ = x ∈ RN : dist(x, ∂Ω) < ρ

where ρ = 12 dist(K, ∂Ω) (> 0). Define v = v(x, t) by

v(x, t) = λ

∫Nρg(x, t; ξ, 0) dξ for every (x, t) ∈ RN × (0,+∞),


where a number λ > 0 will be determined later. Then it follows from (9) that

v(x, t) ≥ λM−1t−N2

∫Nρe−

α|x−ξ|2t dξ for (x, t) ∈ RN × (0,+∞)

and hence we can choose λ > 0 satisfying

v ≥ 1 on ∂Ω× (0, 1].

Thus the comparison principle yields that

u ≤ v in Ω× (0, 1]. (10)

On the other hand, it follows from (9) that

v(x, t) ≤ λMt−N2

∫Nρe−|x−ξ|2αt dξ for (x, t) ∈ RN × (0,+∞).

Since |x− ξ| ≥ ρ for every x ∈ K and ξ ∈ Nρ, we observe that

v(x, t) ≤ λMt−N2 e−

ρ2

αt |Nρ| for every (x, t) ∈ K × (0,+∞),

where |Nρ| denotes the Lebesgue measure of the setNρ. Therefore (10) gives (a).For (c), for instance choose a large ball B with Ω ⊂ B and let ϕ = ϕ(x) be

the first positive eigenfunction of the problem

− div(σ∇ϕ) = λϕ in B and ϕ = 0 on ∂B

with supBϕ = 1. Choose C > 0 sufficiently large to have

1 ≤ Cϕ in Ω.

Then it follows from the comparison principle that

1− u(x, t) ≤ Ce−λtϕ(x) for every (x, t) ∈ Ω× (0,+∞),

which gives (c).

The following asymptotic formula of the heat content of a ball touching at∂Ω at only one point tells us about the interaction between the initial behaviorof solutions and geometry of domain.

Proposition 2.2. Let u be the solution of either problem (1)-(3) or prob-lem (4). Let x ∈ Ω and assume that the open ball Br(x) with radius r > 0


centered at x is contained in Ω and such that Br(x) ∩ ∂Ω = y for somey ∈ ∂Ω. Then we have:

limt→+0

t−N+1

4

∫Br(x)

u(z, t) dz = C(N, σ)

N−1∏j=1

(1

r− κj(y)

)− 1

2

. (11)

Here, κ1(y), . . . , κN−1(y) denote the principal curvatures of ∂Ω at y with respectto the inward normal direction to ∂Ω and C(N, σ) is a positive constant givenby

C(N, σ) =

2σN+1

4s c(N) for problem (1)-(3) ,

2√σm√

σs+√σmσN+1

4s c(N) for problem (4) ,

where c(N) is a positive constant depending only on N . (Notice that if σs = σm

then C(N, σ) = σN+1

4s c(N) for problem (4), that is, just half of the constant for

problem (1)-(3).)When κj(y) = 1/r for some j ∈ 1, · · · , N − 1, (11) holds by setting the

right-hand side to +∞ (notice that κj(y) ≤ 1/r always holds for all j’s).

Proof. For the one-phase problem, that is, for the heat equation ut = ∆u, thislemma has been proved in [12, Theorem 1.1, p. 238] or in [13, Theorem B, pp.2024–2025 and Appendix, pp. 2029–2032]. The proof in [13] was carried outby constructing appropriate super- and subsolutions in a neighborhood of ∂Ωin a short time with the aid of the initial behavior [13, Lemma B.2, p. 2030]obtained by Varadhan’s formula [15] for the heat equation ut = ∆u

−4t log u(x, t)→ dist(x, ∂Ω)2 as t→ +0 (12)

uniformly on every compact set in Ω. (See also [13, Theorem A, p. 2024] forthe formula.) Here, with no need of Varadhan’s formula, (a) of Lemma 2.1gives sufficient information on the initial behavior [13, Lemma B.2, p. 2030].We remark that since problem (1)-(3) is one-phase with conductivity σs near∂Ω, we can obtain formula (11) for problem (1)-(3) only by scaling in t. Onthe other hand, problem (4) is two-phase with conductivities σm, σs near ∂Ω ifσm 6= σs. Therefore, it is enough for us to prove formula (11) for problem (4)where σm 6= σs.

Let u be the solution of problem (4) where σm 6= σs, and let us prove thislemma by modifying the proof of Theorem B in [13, Appendix, pp. 2029–2032].

Let us consider the signed distance function d∗ = d∗(x) of x ∈ RN to theboundary ∂Ω defined by

d∗(x) =

dist(x, ∂Ω) if x ∈ Ω,

− dist(x, ∂Ω) if x 6∈ Ω.(13)


Since ∂Ω is bounded and of class C2, there exists a number ρ0 > 0 such thatd∗(x) is C2-smooth on a compact neighborhood N of the boundary ∂Ω givenby

N = x ∈ RN : −ρ0 ≤ d∗(x) ≤ ρ0. (14)

We make N satisfy N ∩D = ∅. Introduce a function F = F (ξ) for ξ ∈ R by

F (ξ) =1

2√π

∫ ∞ξ

e−s2/4ds.

Then F satisfies

F ′′ +1

2ξF ′ = 0 and F ′ < 0 in R,

F (−∞) = 1, F (0) =1

2, and F (+∞) = 0.

For each ε ∈ (0, 1/4), we define two functions F± = F±(ξ) for ξ ∈ R by

F±(ξ) = F (ξ ∓ 2ε).

Then F± satisfies

F ′′± +1

2ξF ′± = ±εF ′±, F ′± < 0 and F− < F < F+ in R,

F±(−∞) = 1, F±(0) ≷1

2, and F±(+∞) = 0.

By setting η = t−12 d∗(x), µ =

√σm/√σs and θ± = 1 + (µ− 1)F±(0) (> 0),

we introduce two functions v± = v±(x, t) by

v±(x, t) =

µθ±F±

(σ− 1

2s η

)for (x, t) ∈ Ω× (0,+∞),

1θ±

[F±

(σ− 1

2m η

)+ θ± − 1

]for (x, t) ∈ Ωc × (0,+∞).

(15)

Then v± satisfies the transmission conditions

v±∣∣+

= v±∣∣− and σm

∂v±∂ν

∣∣∣+

= σs∂v±∂ν

∣∣∣−

on ∂Ω× (0,+∞), (16)

where + denotes the limit from outside and − that from inside of Ω and ν =ν(x) denotes the outward unit normal vector to ∂Ω at x ∈ ∂Ω, since ν = −∇d∗on ∂Ω. Moreover we observe that for each ε ∈ (0, 1/4), there exists t1,ε ∈ (0, 1]satisfying

(±1) (v±)t − σ∆v± > 0 in (N \ ∂Ω)× (0, t1,ε]. (17)


In fact, a straightforward computation gives

(v±)t − σ∆v± =

− µtθ±

(±ε+

√σst∆d

∗) F ′± in (N ∩ Ω)× (0,+∞),

− 1tθ±

(±ε+

√σmt∆d

∗) F ′± in(N \ Ω

)× (0,+∞).

Then, for each ε ∈ (0, 1/4), by setting t1,ε = 1maxσs,σm

(ε

2M

)2, where M =

maxx∈N|∆d∗(x)|, we obtain (17).

Then, in view of (a) of Lemma 2.1 and the definition (15) of v±, we see thatthere exist two positive constants E1 and E2 satisfying

max|v+|, |v−|, |u| ≤ E1e−E2

t in Ω \ N × (0, 1]. (18)

By setting, for (x, t) ∈ RN × (0,+∞),

w±(x, t) = (1± ε)v±(x, t)± 2E1e−E2

t , (19)

since v± and u are all nonnegative, we obtain from (18) that

w− ≤ u ≤ w+ in Ω \ N × (0, 1]. (20)

Moreover, in view of the facts that F±(−∞) = 1 and F±(+∞) = 0, we see thatthere exists tε ∈ (0, t1,ε] satisfying

w− ≤ u ≤ w+ on ((∂N \ Ω)× (0, tε]) ∪ (N × 0) . (21)

Then, in view of (16), (17), (20), (21) and the definition (19) of w±, we havefrom the comparison principle over N that

w− ≤ u ≤ w+ in(N ∪ Ω

)× (0, tε]. (22)

By writingΓs = x ∈ Ω : d∗(x) = s for s > 0,

let us quote a geometric lemma from [11] adjusted to our situation.

Lemma 2.3. ([11, Lemma 2.1, p. 376]) If max1≤j≤N−1

κj(y) <1

r, then we have:

lims→0+

s−N−1

2 HN−1(Γs ∩Br(x)) = 2N−1

2 ωN−1

N−1∏j=1

(1

r− κj(y)

)− 1

2

,

where HN−1 is the standard (N−1)-dimensional Hausdorff measure, and ωN−1

is the volume of the unit ball in RN−1.


Let us consider the case where max1≤j≤N−1

κj(y) <1

r. Then it follows from (22)

that for every t ∈ (0, tε]

t−N+1

4

∫Br(x)

w− dz ≤ t−N+1

4

∫Br(x)

u dz ≤ t−N+1

4

∫Br(x)

w+ dz. (23)

On the other hand, with the aid of the co-area formula, we have∫Br(x)

v± dz =

µ

θ±(σst)

N+14

∫ 2r(σst)− 1

2

0

F±(ξ)ξN−1

2

((σst)

12 ξ)−N−1

2 HN−1(

Γ(σst)

12 ξ∩Br(x)

)dξ,

where v± is defined by (15). Thus, by Lebesgue’s dominated convergence the-orem and Lemma 2.3, we get

limt→+0

t−N+1

4

∫Br(x)

w± dx =

µ

θ±(σs)

N+14 2

N−12 ωN−1

N−1∏j=1

(1

r− κj(y)

)− 1

2 ∫ ∞0

F±(ξ)ξN−1

2 dξ.

Moreover, again by Lebesgue’s dominated convergence theorem, since

limε→0

θ± = 1 + (µ− 1)F (0) =µ+ 1

2and µ =

√σm/√σs ,

we see that

limt→+0

t−N+1

4

∫Br(x)

w± dx =

2√σm√

σs +√σm

(σs)N+1

4 2N−1

2 ωN−1

N−1∏j=1

(1

r− κj(y)

)− 1

2 ∫ ∞0

F (ξ)ξN−1

2 dξ.

Therefore (23) gives formula (11) provided max1≤j≤N−1

κj(y) <1

r.

Once this is proved, the case where κj(y) = 1/r for some j ∈ 1, · · · , N−1can be dealt with as in [12, p. 248] by choosing a sequence of balls Brk(xk)∞k=1

satisfying:

rk < r, y ∈ ∂Brk(xk), andBrk(xk) ⊂ Br(x) for every k ≥ 1, and limk→∞

rk = r.


Then, because of max1≤j≤N−1

κj(y) ≤ 1

r<

1

rk, applying formula (11) to each ball

Brk(xk) yields that

lim inft→+0

t−N+1

4

∫Br(x)

u(z, t) dz = +∞.

This completes the proof of Proposition 2.2.

In order to determine the symmetry of Ω, we employ the following lemma.

Lemma 2.4. Let u be the solution of either problem (1)-(3) or problem (4).Under the assumption (7) of Theorem 1.1 and Theorem 1.3, the following as-sertions hold:

(a) There exists a number R > 0 such that

dist(x, ∂Ω) = R for every x ∈ Γ;

(b) Γ is a real analytic hypersurface;

(c) there exists a connected component γ of ∂Ω, that is also a real analytichypersurface, such that the mapping γ 3 y 7→ x(y) ≡ y−Rν(y) ∈ Γ, whereν(y) is the outward unit normal vector to ∂Ω at y ∈ γ, is a diffeomorphism;in particular γ and Γ are parallel hypersurfaces at distance R;

(d) it holds that

max1≤j≤N−1

κj(y) <1

Rfor every y ∈ γ, (24)

where κ1(y), · · · , κN−1(y) are the principal curvatures of ∂Ω at y ∈ γ withrespect to the inward unit normal vector −ν(y) to ∂Ω;

(e) there exists a number c > 0 such that

N−1∏j=1

(1

R− κj(y)

)= c for every y ∈ γ. (25)

Proof. First it follows from the assumption (5) that

Br(x) ⊂ Ω \D for every x ∈ ∂G with 0 < r ≤ dist(x, ∂Ω).

Therefore, since σ = σs in Ω\D, we can use a balance law (see [10, Theorem 2.1,pp. 934–935] or [9, Theorem 4, p. 704]) to obtain from (7) that∫

Br(p)

u(z, t) dz =

∫Br(q)

u(z, t) dz for every p, q ∈ Γ and t > 0, (26)


provided 0 < r ≤ min dist(p, ∂Ω), dist(q, ∂Ω). Let us show assertion (a).Suppose that there exist a pair of points p and q satisfying

dist(p, ∂Ω) < dist(q, ∂Ω).

Set r = dist(p, ∂Ω). Then there exists a point y ∈ ∂Ω such that y ∈ Br(p)∩∂Ω.Choose a smaller ball Br(x) ⊂ Br(p) with 0 < r < r and Br(x)∩∂Br(p) = y.Since max

1≤j≤N−1κj(y) ≤ 1

r<

1

r, by applying Proposition 2.2 to the ball Br(x),

we get

lim inft→+0

t−N+1

4

∫Br(p)

u(z, t) dz ≥ limt→+0

t−N+1

4

∫Br(x)

u(z, t) dz > 0.

On the other hand, since Br(q) ⊂ Ω, it follows from (a) of Lemma 2.1 that

limt→+0

t−N+1

4

∫Br(q)

u(z, t) dz = 0,

which contradicts (26), and hence assertion (a) holds true.We can find a point x∗ ∈ Γ and a ball Bρ(z∗) such that Bρ(z∗) ⊂ G and

x∗ ∈ ∂Bρ(z∗). Since Γ satisfies (6), assertion (a) yields that there exists a pointy∗ ∈ ∂Ω satisfying

BR+ρ(z∗) ⊂ Ω, y∗ ∈ BR+ρ(z∗) ∩ ∂Ω, and BR(x∗) ∩ ∂Ω = y∗.

Observe that

max1≤j≤N−1

κj(y∗) ≤1

R+ ρ<

1

Rand x∗ = y∗ −Rν(y∗) ≡ x(y∗).

Define γ ⊂ ∂Ω by

γ =y ∈ ∂Ω : BR(x) ∩ ∂Ω = y for x = y −Rν(y) ∈ Γ

and max1≤j≤N−1

κj(y) <1

R

.

Hence y∗ ∈ γ and γ 6= ∅. By Proposition 2.2 we have that for every y ∈ γ andx = x(y)(= y −Rν(y))

limt→+0

t−N+1

4

∫BR(x)


N−1∏j=1

(1

R− κj(y)

)− 1

2

. (27)

Here let us show that, if y ∈ γ and x = x(y), then ∇u(x, t) 6= 0 for some t > 0,which guarantees that in a neighborhood of x, Γ is a part of a real analytic


hypersurface properly embedded in RN because of (7), real analyticity of u withrespect to the space variables, and the implicit function theorem. Moreover,this together with the implicit function theorem guarantees that γ is open in∂Ω and the mapping γ 3 y 7→ x(y) ∈ Γ is a local diffeomorphism, which isalso real analytic. If we can prove additionally that γ is closed in ∂Ω, thenthe mapping γ 3 y 7→ x(y) ∈ Γ is a diffeomorphism and γ is a connectedcomponent of ∂Ω since Γ is a connected component of ∂G, and hence all theremaining assertions (b) – (e) follow from (26), (27) and the definition of γ.We shall prove this later in the end of the proof of Lemma 2.4.

Before this we show that, if y ∈ γ and x = x(y), then ∇u(x, t) 6= 0 for somet > 0. Suppose that ∇u(x, t) = 0 for every t > 0. Then we use another balancelaw (see [10, Corollary 2.2, pp. 935–936]) to obtain that∫

BR(x)

(z − x)u(z, t) dz = 0 for every t > 0. (28)

On the other hand, (a) of Lemma 2.1 yields that

limt→+0

t−N+1

4

∫K

u(z, t) dz = 0 for every compact set K ⊂ Ω, (29)

and hence by (27) it follows that for every ε > 0

limt→+0

t−N+1

4

∫BR(x)∩Bε(y)


N−1∏j=1

(1

R− κj(y)

)− 1

2

. (30)

This implies that

limt→+0

t−N+1

4

∫BR(x)

(z − x)u(z, t) dz =

C(N, σ)

N−1∏j=1

(1

R− κj(y)

)− 1

2

(y − x) 6= 0,

which contradicts (28).It remains to show that γ is closed in ∂Ω. Let yn be a sequence of points

in γ with limn→∞

yn = y∞ ∈ ∂Ω, and let us prove that y∞ ∈ γ. By combining (26)

with (27), we see that there exists a positive number c satisfying assertion (e)and hence by continuity

N−1∏j=1

(1

R− κj(y∞)

)= c > 0 and max

1≤j≤N−1κj(y

∞) ≤ 1

R, (31)


since yj ∈ γ for every j. Thus max1≤j≤N−1

κj(y∞) <

1

R. Let x∞ = y∞ −

Rν(y∞)(= x(y∞)). It suffices to show that BR(x∞) ∩ ∂Ω = y∞. Supposethat there exists another point y ∈ BR(x∞) ∩ ∂Ω. Then for every R ∈ (0, R)we can find two points p∞ and p in BR(x∞) such that

BR(p∞) ∪BR(p) ⊂ BR(x∞), BR(p∞) ∩ ∂Ω = y∞, and BR(p) ∩ ∂Ω = y.

Hence by Proposition 2.2 we have

limt→+0

t−N+1

4

∫BR(p∞)


N−1∏j=1

(1

R− κj(y∞)

)− 1

2

,

limt→+0

t−N+1

4

∫BR(p)


N−1∏j=1

(1

R− κj(y)

)− 1

2

.

Thus, with the same reasoning as in (30) by choosing small ε > 0, we havefrom (31), (26), (27) and assertion (e) that for every x ∈ γ

C(N, σ)

N−1∏j=1

(1

R− κj(y∞)

)− 1

2

= C(N, σ)c−12

= limt→+0

t−N+1

4

∫BR(x)

u(z, t) dz = limt→+0

t−N+1

4

∫BR(x∞)

u(z, t) dz

≥ limt→+0

t−N+1

4

[∫BR(p∞)∩Bε(y∞)

u(z, t) dz +

∫BR(p)∩Bε(y)

u(z, t) dz

]

= C(N, σ)

N−1∏j=1

(1

R− κj(y∞)

)− 1

2

+

N−1∏j=1

(1

R− κj(y)

)− 1

2

.Since R ∈ (0, R) is arbitrarily chosen, this gives a contradiction, and hence γis closed in ∂Ω.

Lemma 2.5. Let u be the solution of problem (4). Under the assumption (8)of Theorem 1.3, the same assertions (a)–(e) as in Lemma 2.4 hold provided Γand γ are replaced by ∂G and ∂Ω, respectively.

Proof. By the same reasoning as in assertion (a) of Lemma 2.4 we have asser-tion (a) from the assumption (8). Since every component Γ of ∂G has the samedistance R to ∂Ω, every component Γ satisfies the assumption (6). Therefore,


we can use the same arguments as in the proof of Lemma 2.4 to prove thislemma. Here we must have

∂Ω = x ∈ RN : dist(x,G) = R.

3. Proof of Theorem 1.1

Let u be the solution of problem (1)-(3) for N ≥ 2. With the aid of Aleksan-drov’s sphere theorem [1, p. 412], Lemma 2.4 yields that γ and Γ are concentricspheres. Denote by x0 ∈ RN the common center of γ and Γ. By combining theinitial and boundary conditions of problem (1)-(3) and the assumption (7) withthe real analyticity in x of u over Ω \ D, we see that u is radially symmetricwith respect to x0 in x on

(Ω \D

)× (0,∞). Here we used the assumption that

Ω \D is connected. Moreover, in view of the Dirichlet boundary condition (2),we can distinguish the following two cases:

(I) Ω is a ball; (II) Ω is a spherical shell.

By virtue of (c) of Lemma 2.1, we can introduce the following two auxiliaryfunctions U = U(x), V = V (x) by

U(x) =

∫ ∞0

(1− u(x, t)) dt for x ∈ Ω \D, (32)

V (x) =

∫ ∞0

(1− u(x, t)) dt for x ∈ D. (33)

Then we observe that

−∆U =1

σsin Ω \D, −∆V =

1

σcin D, (34)

U = V and σs∂U

∂ν= σc

∂V

∂νon ∂D, (35)

U = 0 on ∂Ω, (36)

where ν = ν(x) denotes the outward unit normal vector to ∂D at x ∈ ∂Dand (35) is the transmission condition. Since U is radially symmetric withrespect to x0, by setting r = |x− x0| for x ∈ Ω \D we have

− ∂2

∂r2U − N − 1

r

∂

∂rU =

1

σsin Ω \D. (37)

Solving this ordinary differential equation yields that

U =

c1r

2−N − 12Nσs

r2 + c2 if N ≥ 3,

−c1 log r − 14σs

r2 + c2 if N = 2,(38)


where c1, c2 are some constants depending on N . Remark that U can be ex-tended as a radially symmetric function of r in RN \ x0.

Let us first show that case (II) does not occur. Set Ω = Bρ+(x0) \Bρ−(x0)

for some numbers ρ+ > ρ− > 0. Since Ω \ D is connected, (36) yields thatU(ρ+) = U(ρ−) = 0 and hence c1 < 0. Moreover we observe that

U ′′ < 0 on [ρ−, ρ+]. (39)

Recall that D may have finitely many connected components. Let us take aconnected component D∗ ⊂ D. Then, since D∗ ⊂ Ω, we see that there existρ∗ ∈ (ρ−, ρ+) and x∗ ∈ ∂D∗ which satisfy

U(ρ∗) = minU(r) : r = |x− x0|, x ∈ ∂D∗ and ρ∗ = |x∗ − x0|. (40)

Notice that ν(x∗) equals either x∗−x0

ρ∗or −x∗−x0

ρ∗. For r > 0, set

U(r) = U(ρ∗) +σsσc

(U(r)− U(ρ∗)). (41)

Since

U(r)− U(r) =

(σsσc− 1

)(U(r)− U(ρ∗)), (42)

it follows that

U

≥ U if σs > σc

≤ U if σs < σcon ∂D∗. (43)

Moreover, we remark that U never equals U identically on ∂D∗ since Ω \D∗ isconnected and Ω is a spherical shell. Observe that

−∆U =1

σcand

∂U

∂r=σsσc

∂U

∂rin D∗. (44)

On the other hand, we have

−∆V =1

σcin D∗ and V = U on ∂D∗. (45)

Then it follows from (43) and the strong comparison principle that

U

> V if σs > σc

< V if σs < σcin D∗, (46)

since U never equals U identically on ∂D∗. The transmission condition (35)with the definition of U tells us that

U = V and∂U

∂ν=∂V

∂νat x = x∗ ∈ ∂D∗, (47)


since ν(x∗) equals either x∗−x0

ρ∗or −x∗−x0

ρ∗. Therefore applying Hopf’s bound-

ary point lemma to the harmonic function U −V gives a contradiction to (47),and hence case (II) never occurs. (See [6, Lemma 3.4, p. 34] for Hopf’s bound-ary point lemma.)

Let us consider case (I). Set Ω = Bρ(x0) for some number ρ > 0. Wedistinguish the following three cases:

(i) c1 = 0; (ii) c1 > 0; (iii) c1 < 0.

We shall show that only case (i) occurs. Let us consider case (i) first. Notethat

U ′(r) < 0 if r > 0, and U ′(0) = 0. (48)

Take an arbitrary component D∗ ⊂ D. Then, since D∗ ⊂ Ω = Bρ(x0), we seethat there exist ρ∗ ∈ (0, ρ) and x∗ ∈ ∂D∗ which also satisfy (40). Notice thatν(x∗) equals x∗−x0

ρ∗. For r ≥ 0, define U = U(r) by (41). Then, by (42) we

also have (43). Observe that both (44) and (45) also hold true. Then it followsfrom (43) and the comparison principle that

U

≥ V if σs > σc

≤ V if σs < σcin D∗. (49)

The transmission condition (35) with the definition of U also yields (47) sinceν(x∗) equals x∗−x0

ρ∗. Therefore, by applying Hopf’s boundary point lemma to

the harmonic function U − V , we conclude from (47) that

U ≡ V in D∗

and hence D∗ must be a ball centered at x0. In conclusion, D itself is connectedand must be a ball centered at x0, since D∗ is an arbitrary component of D.

Next, let us show that case (ii) does not occur. In case (ii) we have

U ′(r) < 0 if r > 0, limr→0

U(r) = +∞, and x0 ∈ D. (50)

Let us choose the connected component D∗ of D satisfying x0 ∈ D∗. Then,since D∗ ⊂ Ω = Bρ(x0), we see that there exist ρ∗1, ρ∗2 ∈ (0, ρ) and x∗1, x∗2 ∈∂D∗ which satisfy that ρ∗1 ≤ ρ∗2 and

U(ρ∗1) = maxU(r) : r = |x− x0|, x ∈ ∂D∗ and ρ∗1 = |x∗1 − x0|, (51)

U(ρ∗2) = minU(r) : r = |x− x0|, x ∈ ∂D∗ and ρ∗2 = |x∗2 − x0|. (52)

Notice that ν(x∗i) equals x∗i−x0

ρ∗ifor i = 1, 2. Also, the case where ρ∗1 = ρ∗2

may occur for instance if D∗ is a ball centered at x0. For r > 0, we set

U(r) =

U(ρ∗2) + σs

σc(U(r)− U(ρ∗2)) if σs > σc ,

U(ρ∗1) + σsσc

(U(r)− U(ρ∗1)) if σs < σc .(53)


Then, as in (43), it follows that

U ≥ U on ∂D∗. (54)

Observe that

−∆U =1

σcand

∂U

∂r=σsσc

∂U

∂rin D∗ \ x0, and lim

x→x0

U = +∞. (55)

Therefore, since we also have (45), it follows from (54) and the strong compar-ison principle that

U > V in D∗ \ x0. (56)

The transmission condition (35) with the definition of U tells us that

U = V and∂U

∂ν=∂V

∂νat x = x∗i ∈ ∂D∗, (57)

since ν(x∗i) equals x∗i−x0

ρ∗ifor i = 1, 2. Therefore applying Hopf’s boundary

point lemma to the harmonic function U −V gives a contradiction to (57), andhence case (ii) never occurs.

It remains to show that case (iii) does not occur. In case (iii), since c1 < 0,there exists a unique critical point r = ρc of U(r) such that

U(ρc) = maxU(r) : r > 0 > 0 and 0 < ρc < ρ ; (58)

U ′(r) < 0 if r > ρc and U ′(r) > 0 if 0 < r < ρc ; (59)

limr→0

U(r) = −∞ and x0 ∈ D. (60)

Let us choose the connected component D∗ of D satisfying x0 ∈ D∗. Then,since D∗ ⊂ Ω = Bρ(x0), as in case (ii), we see that there exist ρ∗1, ρ∗2 ∈ (0, ρ)and x∗1, x∗2 ∈ ∂D∗ which satisfy (51) and (52). In view of the shape of thegraph of U , we have from the transmission condition (35) that at x∗i ∈ ∂D∗, i =1, 2,

∂V

∂ν=σsσc

∂U

∂ν=

0 if ρ∗i = ρc ,

σsσcU ′ if ρ∗i 6= ρc ,

(61)

where, in order to see that ν(x∗i) equals x∗i−x0

ρ∗iif ρ∗i 6= ρc, we used the fact

that both D∗ and Bρ(x0)\D∗ are connected and x0 ∈ D∗. Also, the case whereρ∗1 = ρ∗2 may occur for instance if D∗ is a ball centered at x0. For r > 0, wedefine U = U(r) by

U(r) =

U(ρ∗1) + σs

σc(U(r)− U(ρ∗1)) if σs > σc ,

U(ρ∗2) + σsσc

(U(r)− U(ρ∗2)) if σs < σc .(62)


Remark that (62) is opposite to (53). Then, as in (54), it follows that

U ≤ U on ∂D∗. (63)

Hence, by proceeding with the strong comparison principle as in case (ii), weconclude that

U < V in D∗ \ x0. (64)

Then, it follows from the definition of U and (61) that (57) also holds true. Inconclusion, applying Hopf’s boundary point lemma to the harmonic functionU − V gives a contradiction to (57), and hence case (iii) never occurs.

4. Proof of Theorem 1.3

Let u be the solution of problem (4) for N ≥ 3. For assertion (b) of Theo-rem 1.3, with the aid of Aleksandrov’s sphere theorem [1, p. 412], Lemma 2.4yields that γ and Γ are concentric spheres. Denote by x0 ∈ RN the commoncenter of γ and Γ. By combining the initial condition of problem (4) andthe assumption (7) with the real analyticity in x of u over RN \ D comingfrom σs = σm, we see that u is radially symmetric with respect to x0 in xon(RN \D

)× (0,∞). Here we used the assumption that Ω \D is connected.

Moreover, in view of the initial condition of problem (4), we can distinguishthe following two cases as in section 3:

(I) Ω is a ball; (II) Ω is a spherical shell.

For assertion (a) of Theorem 1.3, with the aid of Aleksandrov’s sphere theorem[1, p. 412], Lemma 2.5 yields that ∂G and ∂Ω are concentric spheres, sinceevery component of ∂Ω is a sphere with the same curvature. Therefore, only thecase (I) remains for assertion (a) of Theorem 1.3. Also, denoting by x0 ∈ RN thecommon center of ∂G and ∂Ω and combining the initial condition of problem (4)and the assumption (8) with the real analyticity in x of u over Ω\D yield thatu is radially symmetric with respect to x0 in x on

(RN \D

)× (0,∞).

By virtue of (b) of Lemma 2.1, since N ≥ 3, we can introduce the followingthree auxiliary functions U = U(x), V = V (x) and W = W (x) by

U(x) =

∫ ∞0

(1− u(x, t)) dt for x ∈ Ω \D, (65)

V (x) =

∫ ∞0

(1− u(x, t)) dt for x ∈ D, (66)

W (x) =

∫ ∞0

(1− u(x, t)) dt for x ∈ RN \ Ω. (67)


Then we observe that

−∆U =1

σsin Ω \D, −∆V =

1

σcin D, −∆W = 0 in RN \ Ω, (68)

U = V and σs∂U

∂ν= σc

∂V

∂νon ∂D, (69)

U = W and σs∂U

∂ν= σm

∂W

∂νon ∂Ω, (70)

lim|x|→∞

W (x) = 0, (71)

where ν = ν(x) denotes the outward unit normal vector to ∂D at x ∈ ∂D or to∂Ω at x ∈ ∂Ω and (69) - (70) are the transmission conditions. Here we used (d)of Lemma 2.1 to obtain (71).

Let us follow the proof of Theorem 1.1. We first show that case (II) forassertion (b) of Theorem 1.3 does not occur. Set Ω = Bρ+(x0) \ Bρ−(x0) forsome numbers ρ+ > ρ− > 0. Since u is radially symmetric with respect to x0 inx on

(RN \D

)× (0,∞), we can obtain from (68)-(71) that for r = |x−x0| ≥ 0

U = c1r2−N − 1

2Nσsr2 + c2 for ρ− ≤ r ≤ ρ+,

W = c3r2−N for r ≥ ρ+,

W = c4 for 0 ≤ r ≤ ρ−,

where c1, . . . , c4 are some constants, since Ω \ D is connected. Remark thatU can be extended as a radially symmetric function of r in RN \ x0. Weobserve that c4 > 0 and c3 > 0. Also it follows from (70) that U ′(ρ−) = 0 andU ′(ρ+) < 0, and hence

c1 < 0 and U ′ < 0 on (ρ−, ρ+].

Then the same argument as in the corresponding case in the proof of Theo-rem 1.1 works and a contradiction to the transmission condition (69) can beobtained. Thus case (II) for assertion (b) of Theorem 1.3 never occurs.

Let us proceed to case (I). Set Ω = Bρ(x0) for some number ρ > 0. Sinceu is radially symmetric with respect to x0 in x on

(RN \D

)× (0,∞), we can

obtain from (68)-(71) that for r = |x− x0| ≥ 0

U = c1r2−N − 1

2Nσsr2 + c2 for x ∈ Ω \D,

W = c3r2−N for r ≥ ρ,

where c1, c2, c3 are some constants, since Ω \D is connected. Remark that Ucan be extended as a radially symmetric function of r in RN \ x0. Thereforeit follows from (70) that U ′(ρ) < 0. As in the proof of Theorem 1.1, Wedistinguish the following three cases:

(i) c1 = 0; (ii) c1 > 0; (iii) c1 < 0.


Because of the fact that U ′(ρ) < 0, the same arguments as in the proof ofTheorem 1.1 works to conclude that only case (i) occurs and D must be a ballcentered at x0.

Acknowledgement.

This research was partially supported by the Grant-in-Aid for ScientificResearch (B) (] 26287020) of Japan Society for the Promotion of Science. Themain results of the present paper were discovered while the author was visitingthe National Center for Theoretical Sciences (NCTS) Mathematics Divisionin the National Tsing Hua University; he wishes to thank NCTS for its kindhospitality.

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Author’s address:

Shigeru SakaguchiResearch Center for Pure and Applied MathematicsGraduate School of Information SciencesTohoku UniversitySendai, 980-8579, JapanE-mail: [email protected]

Received March 13, 2016Revised April 21, 2016

Accepted April 22, 2016

two-phase heat conductors with a stationary isothermic …two-phase heat conductors with a...

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